The problem lies in your lack of knowledge about transistor.
Simply you don't know what saturation is, also in real life the transistor current gain is not constant but it's change with the collector current. And this is why different people use different techniques to find resistor values. And this is what confuse you.

In general, yes. In the case of a BJT the definition is that both the base-emitter and collector-emitter junctions are forward biased. This means the transistor is fully on (as a switch) and the collector-emitter voltage is less than the base-emitter voltage. Typically to insure full saturation, a base current of at least 1/10th of the collector current is used (the transistor Beta value is ignored).

To understand how BJT work, first you need to understand how a constant current source work. Are you familiar with the concept of a current source ?
The BJT's in active region act just like a base current controlled collector current source.
See this example show a BJT in active region.

If base current is flowing the BJT is ON and collector current is BETA (β or Hfe) times larger than the base current (Ic = β*Ib).

And Ic = β *Ib and Ie = Ib + Ic is a basic principle of a transistor "action".

Note that arrow on a BJT (on a emitter) symbol show how current will be flow through BJT. Of course from "+" to "-".

The BJT work very similarly to the tap water valve.
A water valve is always used as a device to control the flow of water. Similarly, always think of a bipolar transistor as a device used to control electric current flow by assistance of a base current.
If base current (Ib) flows hen BJT is ON so there must flow β times Ib current flow through collector. See this picture http://images.elektroda.net/94_1250754403.png

As you can see in first two examples BJT work in active region and Ic current is equal to: Ic = β*Ib = 5mA

In the second example (Ib = 100μA), the transistor is on the edge, between the saturation and active region. Ideally this saturation voltage would be Vce = 0V but in real life it is impossible to happen.

In the last exampleIc = β*Ib = Ic = 100*1mA = 100mA don't holds anymore, do you know why?
Because now we have Rc in the circuit so Ohm's and Kirchoff's law must hold also.
What is the max Ic that can flow in this circuit? We all knows the Ohms law, so Ic_max = Vcc/Rc ≈ 10mA.
BJT tries to create a situation in which the collector current Ic = β*Ib = 100mA. And he lowers his collector-emitter voltage to Vcs_sat voltage. The BJT in full ON. Transistor is in saturation region. And in saturation Ic = β*Ib don't hold anymore. And in saturation the collector is equal to:Ic_sat = (Vcc - Vce_sat)/Rc ≈ Vcc/Rc

And emitter current is always equal to Ib + Ic = 1mA + 10mA = 11mA

Here you have a more detailed explanation about saturation region.

I think a reasonably simple yet useful description of the saturation process goes something like this:

Think of a transistor as a device that tries to create a situation in which the collector current, Ic, is β*Ib. The mechanism by which it does this is by controlling the "resistance" of the collector-emitter path through the device. If the present resistance is such that the collector current is less than β*Ib, then the transistor reduces the resistance. This generally results in a lower collector-emitter voltage which means that the voltage dropped across the external components increases which generally allows more collector current to flow. The reverse happens if the present collector current is more than β*Ib in order to reduce it.

But this process can only be carried so far. The transistor can only lower the collector-emitter voltage so far and, once that point is reached, the transistor no longer has any tools in its bag to increase the collector current up to the point it would like it to be, so the collector-emitter voltage stops dropping and the collector current is less β*Ib.

Ideally this saturation voltage would be 0V and the collector-emitter junction would simply look like a short circuit allowing any current to flow as long as it doesn't go above β*Ib (since, at that point, the transistor will come out of saturation and start increasing Vce in order to hold the current to β*Ib).

In practice, real transistors have a saturation voltage that is greater than 0V, but it is generally only a few hundred millivolts.

written by WBahn

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I hope that know you see that saturation depends only on the Rc value, Vcc and transistor β.

As a small test try to find Ib needed to saturated this transistor in this two circuit.

And I also think I have already spotted that I misunderstood the question in regards to providing the variables to give the result Ib
I was trying to use the Ib's that you provided to get the other variables.. will rework it

If transistor in our circuit will work as a switch, you don't need Hfe from datasheet.
Most BJTs vendors define saturation region when Ic/Ib = 10 (called forced beta) .
And this is why we assume Hfe_sat = 10 http://www.onsemi.com/pub_link/Collateral/P2N2222A-D.PDF (figure 11).
And the most datasheet show Vce_sat for Ic/Ib = 10
So to be one hundred percent sure that your BJT will be in saturation you must use this so-called forced beta technique.
For european BJT Ib/Ib = 20

But if we design an amplifier, we use a minimum hfe value from the data sheet.

For Vcc = 10V; β = 100 and Rc = 100Ω
To saturate this transistor the base current must be larger than
Ib > = (Vcc/Rc)/β = 100mA/100 = 1mA
If we increase Rc resistor to say 500Ω and Ib remains unchanged (1mA).
The transistor will remain in saturation region. The only think that will change is Ic and Ie.
Ic ≈ 10/500Ω ≈ 20mA and Ie ≈ Ib + Ic = 21mA.
And transistor will start leaving the saturation region if the base current is lest than Ib < 20mA/β.

I think I am confused now. Is it true that the larger Ib, the easier it is to make a transistor into saturation?
From the formula, K is only a limited range. Why not K < 10 or something else?
Am I missing something?
For example, if Ic/Ib = 10 (or 20) a transistor will be in saturation and if Ic/Ib < 10 (or 20) is the transistor still in saturation?

This is from an old thread: http://forum.allaboutcircuits.com/showthread.php?p=666864#post666864 I think I am confused now. Is it true that the larger Ib, the easier it is to make a transistor into saturation? From the formula, K is only a limited range. Why not K < 10 or something else? Am I missing something? For example, if Ic/Ib = 10 (or 20) a transistor will be in saturation and if Ic/Ib < 10 (or 20) is the transistor still in saturation?

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I don't understand why you confused?

For this circuit with ideal transistor any base current large than:
Ib > (Vcc/Rc)/β our ideal transistor will be well into saturation.
Do you agree with this statement ?

But in real life ideal transistor don't exist. For any real world transistor the β is not constant. Beta varies with Ic, Vce, temperature. And what is worse, every single transistor will have different beta value and beta will changes for different operating conditions also.
Also in saturation Ic = Ib * β don't work.

To overcome this problem with beta and saturation we are forced to use " overdrive factor"/"Forced Beta" trick.
We simply increase the base current well beyond Ib > (Vcc/Rc)/β (beyond minimum beta). We do this to make sure that we have enough base current to put the transistor well into saturation for every condition we have in our circuit.

For this circuit with ideal transistor any base current large than:
Ib > (Vcc/Rc)/β our ideal transistor will be well into saturation.
Do you agree with this statement ?

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Yes.

But in real life ideal transistor don't exist. For any real world transistor the β is not constant. Beta varies with Ic, Vce, temperature. And what is worse, every single transistor will have different beta value and beta will changes for different operating conditions also.
Also in saturation Ic = Ib * β don't work.

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If so, can we use the formula above with minimum beta, βmin?
If we set Ib > (Vcc/Rc)/βmin will the transistor will be 100% in saturation?
I think there is not true. Otherwise, we don't need to use "Forced Beta" at all.

To overcome this problem with beta and saturation we are forced to use " overdrive factor"/"Forced Beta" trick.
We simply increase the base current well beyond Ib > (Vcc/Rc)/β (beyond minimum beta). We do this to make sure that we have enough base current to put the transistor well into saturation for every condition we have in our circuit.

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Yes, I see that but I am confused about this formula.

Ib = Ic/(βmin * K )
where K = 3...10 - overdrive factor

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If we use Ib > Ic/(βmin * K ) the transistor will be in saturation?
Shouldn't K be less then zero?
Let's consider this example.